Weird Laplace Equation. I'm stuck with this PDE problem. 

Let $R>0$ and let $u:\overline{B(0,R)}\to\mathbb{R}$ be a continuos function such that:
  $$\left\{\begin{matrix} \Delta u +u= u^3 & \text{ in } B(0,R)\\ u = 0 & \text{ on } \partial B(0,R)\end{matrix}\right.$$
  Prove that $|u(x)|\leq 1$ for all $x\in B(0,R)$.

I've managed to prove that:
$$\int_{B(0,R)} (u^4-u^2) \leq 0$$
But it doesn't seem to be useful. (It can be seen by multiplying the equation by $u$ and then integrating using Green's identity).
 A: Observe
\begin{align}
-\int_{B(0, R)} (2k-1) u^{2k-2}|\nabla u|^2=\int_{B(0, R)} u^{2k-1}\Delta u \ dx = \int_{B(0, R)} u^{2k-1}(u^3-u)\ dx \leq 0
\end{align}
which means
\begin{align}
\int_{B(0, R)} u^{2(k+1)}\ dx \leq \int_{B(0, R)} u^{2k}\ dx \leq \ldots \leq \int_{B(0, R)} u^2\ dx
\end{align}
for all $k\geq 1$.
In particular, it follows
\begin{align}
\| u\|_{L^{2(k+1)}(B(0, R))} \leq \| u\|_{L^2(B(0, R))}^{\frac{1}{k+1}}\rightarrow 1
\end{align} 
as $k \rightarrow \infty$. Moreover, since
\begin{align}
\lim_{k\rightarrow \infty}\|u\|_{L^{2(k+1)}(B(0, R))} = \| u\|_{L^\infty(B(0, R))}
\end{align}
then it follows
\begin{align}
\| u\|_{L^\infty(B(0, R))}\leq1.
\end{align}
A: It is not necessary to use $L^p$ estimate. Let $s^+$ define the positive part of $s$, namely if $s\le0$, then $s^+=0$ and if $s>0$, then $s^+=s$. Noting that $(u^2-1)|_{\partial B(0,R)}=0$, by Green's formula, one has
\begin{eqnarray}
\int_{B(0,R)}|\nabla(u^2-1)^+|^2dx&=&\int_{B(0,R)}\nabla(u^2-1)\nabla(u^2-1)^+dx \\
&=&-\int_{B(0,R)}(u^2-1)^+\Delta(u^2-1)dx \\
&=&-\int_{B(0,R)}(u^2-1)^+(2|\nabla u|^2+2u\Delta u)dx\\
&=&-2\int_{B(0,R)}(u^2-1)^+|\nabla u|^2dx-\int_{B(0,R)}(u^2-1)^+u\Delta udx\\
&=&-2\int_{B(0,R)}(u^2-1)^+|\nabla u|^2dx-\int_{B(0,R)}(u^2-1)^+u(u^3-u)dx\\
&=&-2\int_{B(0,R)}(u^2-1)^+|\nabla u|^2dx+\int_{B(0,R)}(u^2-1)^+u^2(1-u^2)dx\\
&\le&0
\end{eqnarray}
So $(u^2-1)^+=0$ or $|u|\le1$.
